134. Gas Station (Array; DP)

標籤：

There are N gas stations along a circular route, where the amount of gas at station i is gas[i].

You have a car with an unlimited gas tank and it costs cost[i] of gas to travel from station i to its next station (i+1). You begin the journey with an empty tank at one of the gas stations.

Return the starting gas station‘s index if you can travel around the circuit once, otherwise return -1.

Note:
The solution is guaranteed to be unique.

思路：

1. 是否能travel around? =>只要gas[]加總的和 - cost[]加總的和 > 0，就能夠。=>需要一個INT計算gas[]-cost[]的總合

2. 如何找到起點？如果到了某一站油不夠，那麼起點必須從它後面開始重新找。=>需要一個INT記錄從找到的起點到目前為止的油量

時間複雜度O(n)

class Solution {public:    int canCompleteCircuit(vector<int> &gas, vector<int> &cost) {        int sum = 0, total = 0, len = gas.size(), index = -1;        for(int i=0; i<len; i++){            sum += gas[i]-cost[i];            total += gas[i]-cost[i];            if(sum < 0){                index = i;                sum = 0;            }        }        return total>=0 ? index+1 : -1;    }};

 

134. Gas Station (Array; DP)